Basic Terms In Statistics

Just before I start with detailing some basic terms in statistics, I would emphasise that I am not a statistician, however, anticipate that I have presented enough details below to clarify particular terms. This article details a couple of simple statistics and attempts to expand these with some comments on just how they may be used in practice. The use of statistics in the management of risk is enormous and it is not actually the intention below to produce authentic examples in this field but to make certain that you get some notion of their essential meaning.

Mean (expected value or average):

We could take any kind of activity and measure values that illustrate that event. Let us say that we measure 12 values, they could be:

The value of 9.25 is the 'expected value' which you may anticipate through determining a one off value for the activity, but of course, in practice there is a range of values. The 'expected values' are accumulative.

Supposing that we had 4 individual events and we took similar measurements for each one we might finish up having 'expected values' for each activity of:

5.3, 8.4, 6.3 and 11.3

If we next took into account the result of determining the outcome of the 4 activities we might just predict an overall value of:

(5.3 + 8.4 + 6.3 + 11.3) = 31.3

Once again, the true value will be in a range.

Likelihood:

We referred above to an activity where we attained the following values:

Values: 3, 5, 5, 6, 7, 9, 10, 11, 12, 12, 15, 16

If we were to anticipate an activity with these values we would be assuming that they are each equally likely for the calculation of the 'expected value' or the mean. Simply put if there is little possibility of the initial 6 values being measured, then just the last 6 matter, so the 'expected value' will be:

(10 + 11 + 12 + 12 + 15 + 16) / 6 = 76/6 = 12.7

Thus, in practice not only is there a distribution of values, their likelihood of occurring may equally be different.

We may have a basic look at a task in a project. We may want to know how long it could be postponed before it begins. A project manager may ask the expert for his opinion and he might advocate 16 weeks. The project manager can take advantage of this information concerning his planning. However, this estimation is going to be based upon certain assumptions which the project manager ought to challenge.

If the expert is 100 % certain that there will be 16 working weeks delay that's good, but it is not usually the situation. What we do realize is that there will be a setback. The probability of this coming about is 1, that is,, it will occur. We can, at present, take into account other potential scenarios.

If we suppose that we possess a range of hold ups (in weeks) each with a particular likelihood. We could get:

(Delay)..........(Probability)..........(Contribution)

6.........................0.3......................6 x 0.3 = 1.8

16.......................0.5.....................16 x 0.5 = 8.0

20.......................0.2.....................20 x 0.2 = 4.0

The 'contribution' represents a 'weighted' value. Note that the sum of each one of the chances adds up to 1, which must be the probability of the hold up actually occurring.

The expected value this time becomes:

(1.8 + 8.0 + 4.0) = 13.8 weeks

This is the more probable value that the project manager could make use of in his plans rather than 16 working weeks.

If we considered a selection of 4 comparable activities we may finish up with an overall hold up of 13.8 x 4 = 55.2 working weeks (if executed in series and not in parallel). If we had taken a single estimate of 16 working weeks the potential complete hold up to the project would have come to 64 weeks (approximately 16 % longer).

When evaluating a single activity the effect is not much of a problem, but when evaluating several events the differences can certainly accumulate.

The value of 16 working weeks hold up becomes the most likely (highest chance) and is 2.2 weeks more than the 13.8 working weeks above.

It is due to the fact that the spread of the values is 'skewed' a little. Had there been a balanced distribution the 'expected value' would have worked out as being the same as the initial estimation, that is, 16 weeks.

(Delay)..........(Probability)..........(Contribution)

12......................0.25......................12 x 0.25 = 3.0

16......................0.5........................16 x 0.5 = 8.0

20......................0.25.......................20 x 0.2 = 5.0

Here the contributions are:

(3.0 + 8.0 + 5.0) = 16 weeks

Ideally, the above has given a small awareness into one of the basic terms in statistics, the 'mean' of a group of values'.

There will be a few more in the next article.

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